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  操作系统-11-经典进程同步互斥问题
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      <h1 id="操作系统-11-经典进程同步互斥问题"><a href="#操作系统-11-经典进程同步互斥问题" class="headerlink" title="操作系统-11-经典进程同步互斥问题"></a>操作系统-11-经典进程同步互斥问题</h1><h2 id="1-消费者-生产者问题"><a href="#1-消费者-生产者问题" class="headerlink" title="1. 消费者-生产者问题"></a>1. 消费者-生产者问题</h2><p><strong>1. 问题描述</strong></p>
<ul>
<li>系统中有<code>一组生产者进程</code>和<code>一组消费者进程</code>，生产者进程每次<code>生产一个</code>产品放入缓冲区，消费者进程每次从缓冲区中<code>取出一个</code>产品并使用。(注: 这里的“产品”理解为某种数据)</li>
<li>生产者、消费者<code>共享</code>一个初始为空、大小为n的<code>缓冲区</code>。</li>
<li>只有缓冲区<code>没满</code>时，<code>生产者</code>才能把产品<code>放入</code>缓冲区，否则必须等待。</li>
<li>只有缓冲区<code>不空</code>时，<code>消费者</code>才能从中<code>取出</code>产品，否则必须等待。</li>
<li>缓冲区是临界资源，各进程必须<code>互斥</code>地访问。</li>
</ul>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201009/091536548.png" alt="mark"></p>
<a id="more"></a>



<ol start="2">
<li><strong>问题分析</strong></li>
</ol>
<ul>
<li><strong>1)关系分析</strong>。生产者和消费者对缓冲区互斥访问是<code>互斥关系</code>，同时生产者和消费者又是一个相互协作的关系，只有生产者生产之后,消费者才能消费，它们也是<code>同步关系</code>。</li>
<li><strong>2)整理思路</strong>。根据各进程的操作流程确定P、V操作的大致顺序。<br>生产者每次要消耗(P）一个空闲缓冲区，并生产(V)一个产品。<br>消费者每次要消耗(P）一个产品，并释放一个空闲缓冲区(V)。<br>往缓冲区放入/取走产品需要互斥。</li>
<li><strong>3)信号量设置。</strong>设置信号量。设置需要的信号量，并根据题目条件确定信号量初值。( 互斥信号量初值一般为1，同步信号量的初始值要看对应资源的初始值是多少)</li>
</ul>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201009/091728863.png" alt="mark"></p>
<ol start="3">
<li><strong>实现</strong></li>
</ol>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201009/091816753.png" alt="mark"></p>
<p><strong>注意 ： 实现互斥的P操作一定要在实现同步的P操作之后（否则会有死锁问题）</strong></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201009/092337577.png" alt="mark"></p>
<p><strong>单生产者-消费者回顾</strong></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201009/092555328.png" alt="mark"></p>
<h2 id="2-多生产者-消费者问题"><a href="#2-多生产者-消费者问题" class="headerlink" title="2. 多生产者-消费者问题"></a>2. 多生产者-消费者问题</h2><ol>
<li><strong>问题描述</strong></li>
</ol>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/142657247.png" alt="mark"></p>
<ol start="2">
<li><strong>问题分析</strong></li>
</ol>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/142750028.png" alt="mark"></p>
<ol start="3">
<li><strong>实现方式</strong></li>
</ol>
<p><strong>① 有mutex</strong></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/142951274.png" alt="mark"></p>
<p><strong>② 无mutex</strong></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/143000426.png" alt="mark"></p>
<p><strong>为什么有mutex和没有mutex一样呢？</strong></p>
<ul>
<li>原因在于:本题中的缓冲区大小为1，在任何时刻，apple、 orange、 plate 三个同步信号量中最多只有一个是1。</li>
<li>因此在任何时刻，最多只有一个进程的P操作不会被阻塞，并顺利地进入临界区…</li>
</ul>
<p><strong>如果有两个盘子plate</strong></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/143354331.png" alt="mark"></p>
<ol start="4">
<li><h5 id="知识总结与重要考点"><a href="#知识总结与重要考点" class="headerlink" title="知识总结与重要考点"></a>知识总结与重要考点</h5></li>
</ol>
<p><code>总结</code>:在生产者_消费者问题中，如果缓冲区大小为1，那么有可能不需要设置互斥信号量就可以实现互斥访问缓冲区的功能。当然，<code>这不是绝对的</code>，要具体问题具体分析。</p>
<p><code>建议</code>:在考试中如果来不及仔细分析，可以加上互斥信号量，保证各进程一定会互斥地访问缓冲区。但需要注意的是，·<code>实现互斥的P操作一定要在实现同步的P操作之后</code>·，否则可能引起·<code>“死锁”</code>·。</p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/143729758.png" alt="mark"></p>
<h2 id="3-吸烟者问题"><a href="#3-吸烟者问题" class="headerlink" title="3. 吸烟者问题"></a>3. 吸烟者问题</h2><ol>
<li><strong>问题描述</strong></li>
</ol>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201009/092714018.png" alt="mark"></p>
<ol start="2">
<li><h5 id="问题分析"><a href="#问题分析" class="headerlink" title="问题分析"></a>问题分析</h5></li>
</ol>
<ul>
<li><strong>本质是单生产者 - 多消费者的问题</strong></li>
</ul>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201009/093058018.png" alt="mark"></p>
<ol start="3">
<li><strong>实现方法</strong></li>
</ol>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201009/093158353.png" alt="mark"></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201009/093214151.png" alt="mark"></p>
<p><strong>回顾 :</strong> </p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201009/093805017.png" alt="mark"></p>
<h2 id="4-读者写者问题"><a href="#4-读者写者问题" class="headerlink" title="4. 读者写者问题"></a>4. 读者写者问题</h2><ol>
<li><strong>问题描述</strong></li>
</ol>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/143934790.png" alt="mark"></p>
<ol start="2">
<li><strong>问题分析</strong></li>
</ol>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/144022559.png" alt="mark"></p>
<ol start="3">
<li><h5 id="实现方法"><a href="#实现方法" class="headerlink" title="实现方法"></a>实现方法</h5></li>
</ol>
<p><strong>① 给count加mutex互斥访问</strong></p>
<ul>
<li>这里说一下为什么要加mutex。</li>
<li>比如：当count=0时，第一个读者进程执行到p(rw),rw=0,假设此时时间片到了，切换到第二个读者进程,第二个进程发现count=0,则执行p(rw)，但是此时rw=0，于是第二个进程被堵在p（rw）这里，同理，后面的可能会有多个进程堵在p(rw)，只有当第一个进程再次获得时间片，执行count++,让count不为0，然后其他进程就可以直接绕过if直接进行count++来访问文件，但是第三个读者进程和后面的几个可能堵在p(rw)的多个读者进程则必须得等count–为0后才可以再次和写进程竞争来访问文件，对count的访问没有做到一气呵成，会导致本来一些进程一直堵在p（rw）。</li>
</ul>
<p><strong>② 加一个w实现“读写公平法”</strong></p>
<ul>
<li>在上面的算法中，读进程是优先的，即当存在读进程时，写操作将被延迟，且只要有 一个读进程活跃，随后而来的读进程都将被允许访问文件。这样的方式会导致写进程可能长时间等待，且存在写进程<code>“饿死”</code>的情况。</li>
<li>若希望写进程优先，<code>即当有读进程正在读共享文件时，有写进程请求访问，这时应禁止后续读进程的请求，等到已在共享文件的读进程执行完毕，立即让写进程执行，只有在无写进程执行的情况下才允许读进程再次运行</code>。</li>
<li>为此，增加一个信号量并在上面程序的<code>writer()</code>和 <code>reader()</code>函数中各增加一对PV操作，就可以得到写进程优先的解决程序。</li>
</ul>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/145521452.png" alt="mark"></p>
<ol start="4">
<li><h5 id="知识回顾与重要考点"><a href="#知识回顾与重要考点" class="headerlink" title="知识回顾与重要考点"></a>知识回顾与重要考点</h5></li>
</ol>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/145551296.png" alt="mark"></p>
<h2 id="5-哲学家进餐问题"><a href="#5-哲学家进餐问题" class="headerlink" title="5. 哲学家进餐问题"></a>5. 哲学家进餐问题</h2><ol>
<li><h5 id="问题描述"><a href="#问题描述" class="headerlink" title="问题描述"></a>问题描述</h5></li>
</ol>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/145634639.png" alt="mark"></p>
<ol start="2">
<li><h5 id="问题分析-1"><a href="#问题分析-1" class="headerlink" title="问题分析"></a>问题分析</h5></li>
</ol>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/145759644.png" alt="mark"></p>
<ol start="3">
<li><h5 id="如何实现"><a href="#如何实现" class="headerlink" title="如何实现"></a>如何实现</h5></li>
</ol>
<ul>
<li><strong>实现一：最多允许四个哲学家同时进餐</strong></li>
<li><strong>实现二：</strong><ul>
<li><strong>奇数号哲学家先拿左边的筷子，再拿右边的筷子</strong></li>
<li><strong>偶数号哲学家先拿右边的筷子，再拿左边的筷子</strong></li>
</ul>
</li>
<li><strong>实现三：仅当一名哲学家左右两边筷子都可用的时候才允许拿起筷子</strong></li>
</ul>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/150150502.png" alt="mark"></p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/150153963.png" alt="mark"></p>
<p><strong>例子</strong></p>
<p><strong>实现三：仅当一名哲学家左右两边筷子都可用的时候才允许拿起筷子</strong> 为例</p>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/150440432.png" alt="mark"></p>
<ol start="4">
<li><strong>总结和回顾</strong></li>
</ol>
<p><img src="http://zhuuu-bucket.oss-cn-beijing.aliyuncs.com/img/20201017/150635987.png" alt="mark"></p>
<p><strong>参考书籍：《王道考研计算机操作系统》</strong></p>
<p><strong>参考博客</strong> ：<a href="https://mubu.com/doc/Cd-Y4YOfkh#o-1d01735b78f18c07f" target="_blank" rel="noopener">https://mubu.com/doc/Cd-Y4YOfkh#o-1d01735b78f18c07f</a></p>

      
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